Local Homeomorphism with constant fiber is a covering map

Let $$f:X\to Y$$ be a local homeomorphism between Hausdorff spaces such that $$f$$ is surjective and $$f^{-1}(y)$$ is compact $$\forall y\in Y$$. If $$|f^{-1}(y_1)|=|f^{-1}(y_2)|,\forall y_1,y_2\in Y$$ then show that $$f$$ is a covering map.

Attempt: I know $$f^{-1}(y)$$ is finite as it has a finite cover $$\cup_{i=1}^nU_i\supseteq f^{-1}(y)$$ with each $$U_i$$ open such that $$f_{U_i}$$ is a homeomorphism. I considered $$V_i\subseteq U_i$$ with $$V_i\cap V_j=\emptyset$$ but the problem is the preimage of $$\cap_{i=1}^nf(V_i)$$ is not necessarily in $$\cup_{i=1}^nU_i$$ otherwise I would have found an evenly covered neighborhood of $$y$$. I need to show that $$f$$ is closed so I can use the set $$(\cap_{i=1}^nf(V_i))\backslash (f(X\backslash \cup_{i=1}^nU_i))$$ instead or consider another approach by using the constant preimage assumption which I do not see how it fits there...

It is much simpler than I thought it seems: the sets $$V_i$$ are the disjoint restrictions of the $$U_i$$ as defined in my question above. Suppose that $$|f^{-1}(y)|=n,\forall y\in Y$$. The problem is that taking the set $$V=\cup_{i=1}^nf(V_i)$$, I want to show that $$f^{-1}(V)\subseteq \cup_{i=1}^nV_i$$ so I can use the local 1-1 property and conclude that the preimages are exactly the open sets $$f^{-1}(V)\cap U_i$$. This is straghtforward however: if $$y_0\in f^{-1}(V)$$ then there exist $$v_1\in V_1,...,v_n\in V_n: f(v_i)=y_0$$ as we simply have $$v_i=f_{V_i}^{-1}(y_0)$$. But $$|f^{-1}(y_0)|=n \implies f^{-1}(y_0)=\{v_1,...,v_n\}$$ and thus $$f^{-1}(y_0)\subseteq \cup_{i=1}^nV_i, \forall y_0\in Y$$. The result now follows immediately.

• Yes, that seem correct. I guess the suggestion to prove the map closed led us all astray!
– Ruy
Commented Dec 30, 2020 at 18:34
• Indeed! We all miss some trivial results and take a longer route sometimes... Commented Dec 30, 2020 at 18:39

Apparently the only remaining difficulty is to show that $$f$$ is closed, so let us concentrate in proving this.

So let $$C$$ be a closed subset of $$X$$ and pick $$y\in \overline{f(C)}.$$ Write $$f^{-1}(y)=\{x_1, x_2, \ldots , x_n\},$$ and observe that $$n$$ is therefore the constant number of elements of each fiber of $$f$$.

For each $$k=1,\ldots ,n$$, choose open sets $$U_k\subseteq X$$ and $$V_k\subseteq Y$$, such that $$x_k\in U_k$$, $$y\in V_k$$, and $$f$$ is a homeomorphism from $$U_k$$ onto $$V_k$$. By using that $$X$$ is Hausdorff and shrinking the $$U_k$$ appropriately we may assume that the $$U_k$$ are pairwise disjoint.

Without loss of generality we may also assume that the $$V_k$$ are all the same since otherwise we may replace each $$V_k$$ by $$V=\bigcap_{k=1}^n V_k,$$ while replacing each $$U_k$$ by $$f^{-1}(V)\cap U_k$$.

Since $$y\in \overline{f(C)}$$, we may find a net $$\{c_i\}_{i\in I}\subseteq C$$, such that $$y=\lim_{i\in I}f(c_i)$$. By discarding an initial segment of our net we may assume that $$f(c_i)\in V$$, for all $$i$$, so we may choose $$d^k_i\in U_k$$, such that $$f(d^k_i)=f(c_i)$$.

Due to the fact that the $$U_k$$ are pairwise disjoint we have that, for each $$i$$, the elements $$d^1_i, d^2_i,\ldots , d^n_i$$ are all distinct, so there are precisely $$n$$ of them, and since they all lie in $$f^{-1}\big (f(c_i)\big )$$ we deduce from the hypothesis that $$f^{-1}\big (f(c_i)\big ) =\{d^1_i, d^2_i,\ldots , d^n_i\}.$$

It follows that $$c_i\in \{d^1_i, d^2_i,\ldots , d^n_i\},$$ for every $$i$$, so the pigeonhole principle implies that there is a $$k$$ and a co-final subset $$I_0\subseteq I$$, such that $$c_i=d^k_i$$ for all $$i\in I_0$$. We then have that $$x_k = \lim_{i\in I}d^k_i = \lim_{i\in I_0}d^k_i = \lim_{i\in I_0}c_i \in C,$$ so $$y=f(x_k)\in f(C),$$ whence $$f(C)$$ is closed.

• I cannot find anything wrong so far, seems like a nice approach. That would imply that I do not need the same number of elements in each fiber but having all fibers $|f^{-1}(y)|\leq M$ for a specified number $M$ as pigeonhole principle would give me an $m\leq M$ and a cofinal subset $I_1\subseteq I$ with $c_i\in\{d_i^1,...,d_i^m\},\forall i\in I_1$ right? Commented Dec 30, 2020 at 17:15
• I don't think a bound on the size of each inverse image would suffice. A counter example would be to take $X$ the disjoint union of $\mathbb R$ and $(0,\infty )$, $Y=\mathbb R$, and $f$ the identity map on each component of $X$. That the inverse images are all of the same size is also used in my argument leading up to $$f^{-1}\big (f(c_i)\big ) =\{d^1_i, d^2_i,\ldots , d^n_i\}.$$
– Ruy
Commented Dec 30, 2020 at 17:38
• I think one may get away with assuming that the function $$y↦|f^{-1}(y)|$$ is upper semicontinuous.
– Ruy
Commented Dec 30, 2020 at 17:42
• Of course! Your argument is correct as far as I can tell, I upvoted it but won't accept because there is a much more direct approach I should have seen. Your answer helped me realize this Commented Dec 30, 2020 at 18:24

You don't need to show that $$f$$ is closed.

Let $$y \in Y$$. Since the fiber $$f^{-1}(y)$$ is compact, then it is closed because $$X$$ is Hausdorff, also, since $$f^{-1}(y)$$ is finite, then it is discrete.

Say $$f^{-1}(y) = \{x_1, x_2, \dots, x_n\}$$, then by the Hausdorff property of $$X$$, we can find open sets $$\{U_1, U_2, \dots, U_n\}$$ such that $$x_i \in U_i$$ and $$U_i \cap U_j = \phi$$ for each $$i \neq j$$.

Since $$f$$ is a local homeomorphism, we can find for each $$x_i$$ an open neighborhood $$V_i$$ such that $$f(V_i)$$ is an open neighborhood of $$y$$ and that the restriction $$f|_{V_i}$$ is a homeomorphism.

This implies that $$W_i = U_i\cap V_i$$ is an open neighborhood of $$x_i$$ for all $$i$$ and that $$f(W_i)$$ is an open neighborhood of $$y$$ from the fact that $$f|_{V_i}$$ is a homeomorphism, and thus, maps open sets in the subspace $$V_i$$ (each $$W_i$$ is an open set in the subspace $$V_i$$) to open sets in the subspace $$f(V_i)$$ which are also open sets in $$Y$$. Also, each $$f|_{W_i}$$ is a homeomorphism.

Therefore, $$B = \bigcap\limits_{i}^{n}f(W_{i})$$ is an open neighborhood of $$y$$.

Moreover, let $$A_i = f^{-1}|_{W_i}(B) = W_i \cap f^{-1}(B)$$. Since $$f|_{W_i}$$ is a homeomorphism onto $$f(W_i)$$, then it is also a surjection and an injection, and from the fact that $$A_i \subseteq W_i$$, $$f(A_i) = f|_{W_i}(A_i) = B$$. So each $$A_i$$ is mapped homeomoprhically to $$B$$.

Also, $$x_i \in A_i$$ for all $$i$$, and thus, $$f^{-1}(y) \subseteq \bigcup\limits_{i}^{n}A_i$$. Hence, $$\{y\} = f(f^{-1}(y)) \subseteq f(\bigcup\limits_{i}^{n}A_i) = \bigcup\limits_{i}^{n}f(A_i) = B$$. In other words, $$B$$ is a neighborhood of $$y$$.

The problem now is that we are not sure whether $$f^{-1}(B)$$ is contained in $$\bigcup\limits_{i}^{n}A_i$$.

Let $$x' \in f^{-1}(B)$$, then $$f(x') = y' \in B$$. Therefore, $$y' \in f(W_i)$$ for all $$i$$. Hence, we can find $$x'_i$$ in each $$W_i$$ such that $$f(x'_i) = y'$$. Using the fact that $$|f^{-1}(y_1)| = |f^{-1}(y_2)|$$ for all $$y_1 \neq y_2$$ and that $$|f^{-1}(y)| = n$$, we conclude that $$f^{-1}(y')=\{x'_1, x'_2, \dots, x'_n\} \subseteq \bigcup\limits_{i}^{n}A_i$$. But $$x'$$ must be one of $$\{x'_1, x'_2, \dots, x'_n\}$$, and thus, $$x' \in \bigcup\limits_{i}^{n}A_i$$ which completes the proof.